OK, I made some simplifying assumptions. I hope they are not wrong.

Assumption 2: House interior allowed to cool to 50 F until 4:00pm. Then is increased to 70F over one hour. Assume (nearly) constant drop of interior temps over the day.

OK…let’s assume that 1/10 of the volume of the house is wood, and 9/10 air.

Total house volume is therefore 30*40*12 = 14,400 ft[sup]3[/sup]

Wood, at 36 lbm/ft[sup]3[/sup], makes up 1/30 of that, or 480 ft[sup]3[/sup]. This equates to 17,280 lbm wood.

Air, at 0.063 lbm/ft[sup]3[/sup], makes up 29/30 of that, or 13,920 ft[sup]3[/sup]. This equates to 878 lbm air.

At specific heat capacities of 0.4 Btu/lbm*R for wood and 0.24 Btu/lbm*R for air, we have a house heat capacity of:

17,280 lbm * 0.4 Btu/lbm*R + 878 lbm * 0.24 Btu/lbm*R = 7123 Btu/R.

Map: Exterior temperatures / Interior Temperatures

Since I’m not going to do the differential equation, I’ll guesstimate. I’ll assume that the temperature drop of the house is assessed at the end of each hour.

7-8am: Energy = 2880 ft[sup]2[/sup]*(70-21.25)/19 = 7389.5 Btu

New house temperature: 70 - (7389.5 / 7123) = 68.96 F

8-9am: Energy = 2880*(68.96-23.75)/19 = 6853 Btu

New house temperature: 68.96 - (6853 / 7123) = 68 F

9-10am: Energy = 2880*(68-26.25)/19 = 6328 Btu

New house temperature: 68 - (6328 / 7123) = 67.1 F

10-11am: Energy = 2880*(67.1-28.75)/19 = 5813 Btu

New house temperature: 67.1 - (5813 / 7123) = 66.3 F

11-12pm: Energy = 2880*(66.3-31.25)/19 = 5310 Btu

New house temperature: 66.3 - (5310 / 7123) = 65.6 F

12-1pm: Energy = 2880*(65.6-33.75)/19 = 4821 Btu

New house temperature: 65.6 - (4821 / 7123) = 64.9 F

1-2pm: Energy = 2880*(64.9-36.25)/19 = 4346 Btu

New house temperature: 64.9 - (4346 / 7123) = 64.3 F

2-3pm: Energy = 2880*(64.3-38.75)/19 = 3871 Btu

New house temperature: 64.3 - (3871 / 7123) = 63.75 F

3-4pm: Energy = 2880*(63.75-37.5)/19 = 3979 Btu

New house temperature: 63.75 - (3979 / 7123) = 63.2 F

Now: It’s 4:00pm. We need to heat the house to 70 F. This means we need to regain the heat we lost, and account for continued heat losses over the hour. First - the heat losses over the hour of heating:

4-5pm: Energy = 2880*(((70+63.2)/2) - 32.5)/19 = 5168 Btu

Now, add to that the heat required to raise the house from 63.2 F to 70 F:

(70-63.2)*7123 = 48,436 Btu

Adding the two gives us:

48,436 + 5168 = 53,604 Btu

Which is less heat required than the previous example.

Please check for math and logic errors. I’m in a hurry, and may have screwed up.